What is the molarity of an acetic acid (CH3COOH) if 27.5 mL of 0.25 M NaOH are needed to neutralize 16.0 mL of the sample? M CH3COOH

Answer 1

Acid/base neutralization reaction:

CH3COOH + NaOH ==L CH3COONa + H2O

moles NaOH needed = 27.5 ml NaOH x 1 L/1000 ml x 0.25 mol/L = 0.006875 moles NaOH

moles CH3COOH present = 0.006875 mol NaOH x 1 mol CH3COOH/mol NaOH = 0.006875 mol CH3COOH

Molarity of CH3COOH = mol/liter

moles = 0.006875

liters = 16.0 ml x 1 L/1000 ml = 0.0160 L

M = 0.006875 mol/0.0160 L = 0.43 M (to 2 significant figures)